Medium MCQ +4 / -1 PYQ · JEE Mains 2020

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperature, T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2 . T is temperature of the sink of first engine which is also the source for the second which is also the source for the second engine. How is T related to T1 and T2 . If both engines perform equal amount of work?

  1. A $T = {{2{T_1}{T_2}} \over {{T_1} + {T_2}}}$
  2. B $T = \sqrt {{T_1}{T_2}}$
  3. C $T = {{{T_1} + {T_2}} \over 2}$ Correct answer
  4. D T = 0

Solution

Q<sub>H</sub> : Heat input to 1<sup>st</sup> engine <br><br>Q<sub>L</sub> : Heat rejected from 1<sup>st</sup> engine <br><br>Q'<sub>L</sub> : Heat rejected from 2<sup>nd</sup> engine <br><br>T : Lower temperature of first engine <br><br>Work done by 1<sup>st</sup> engine = work done by 2<sup>nd</sup> engine <br><br>Q<sub>H</sub> – Q<sub>L</sub> = Q<sub>L</sub> – Q'<sub>L</sub> <br><br>$\Rightarrow$ 2Q<sub>L</sub> = Q<sub>H</sub> + Q'<sub>L</sub> <br><br>$\Rightarrow$ 2 = ${{{Q_H}} \over {{Q_L}}} + {{Q{'_L}} \over {{Q_L}}}$ <br><br>$\Rightarrow$ 2 = ${{{T_1}} \over T} + {{{T_2}} \over T}$ <br><br>$\Rightarrow$ $T = {{{T_1} + {T_2}} \over 2}$

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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